Đáp án:
Ta có:
$\begin{array}{l}
+ )\sin C = \frac{{AB}}{{BC}}\\
\Rightarrow AB = sinC.BC = \sin {30^0}.10 = \frac{1}{2}.10 = 5\left( {cm} \right)\\
+ )\cos C = \frac{{AC}}{{BC}}\\
\Rightarrow AC = \cos C.BC = \cos {30^0}.10 = \frac{{\sqrt 3 }}{2}.10 = 5\sqrt 3 \left( {cm} \right)\\
+ ){S_{ABC}} = \frac{1}{2}.AB.AC = \frac{1}{2}.AH.BC\\
\Rightarrow AH = \frac{{AB.AC}}{{BC}} = \frac{{5.5\sqrt 3 }}{{10}} = \frac{{5\sqrt 3 }}{2}\left( {cm} \right)\\
+ )A{B^2} = BH.BC\\
\Rightarrow BH = \frac{{A{B^2}}}{{BC}} = \frac{{{5^2}}}{{10}} = 2,5\left( {cm} \right)
\end{array}$
(Theo hệ thức lượng trong tam giác)
